SOLUTION: How many gallons of acid must be mixed with 40 gallons of 60% acid solution to get a 75% solution?

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Question 180957This question is from textbook College Algebra
: How many gallons of acid must be mixed with 40 gallons of 60% acid solution to get a 75% solution? This question is from textbook College Algebra

Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
Let x= amount of pure acid that is needed
Now we know that the amount of pure acid that is needed(x) plus the amount of pure acid in the 60% solution(0.60*40) has to equal the amount of pure acid in the final mixture(0.75(40+x)). So our equation to solve is:
x+0.60*40=0.75(40+x) get rid of parens and simplify
x+24=30+0.75x subtract 0.75x and also 24 from each side
x-0.75x+24-24=30-24+0.75x-0.75x collect like terms
0.25x=6 divide each side by 0.25
x=24 gal-------------------------------amount of pure acid needed
CK
24+0.60*40=0.75*64
24+24=48
48=48
Hope this helps---ptaylor